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Question Number 29159 by abdo imad last updated on 04/Feb/18

l e t g i v e f (x) = 2 \sqrt{x - 1} + 3 x f i n d f^{- 1} (x) a n d {(f^{- 1})}^{^{'}} (x) .

Commented by abdo imad last updated on 08/Feb/18

f (x) = y \Leftrightarrow x = f^{- 1} (y) \Leftrightarrow 2 \sqrt{x - 1} + 3 x = y

\Leftrightarrow 2 \sqrt{x - 1} = y - 3 x \Leftrightarrow 4 (x - 1) = {(y - 3 x)}^{2} w i t h t h e c o n d .

x - 1 ⩾ 0 a n d y - 3 x ⩾ 0 \Leftrightarrow 4 x - 4 = y^{2} - 6 x y + 9 x^{2}

\Leftrightarrow 9 x^{2} - 6 x y - 4 x + y^{2} + 4 = 0

\Leftrightarrow 9 x^{2} - 2 (3 y + 2) x + y^{2} + 4 = 0

Δ^{^{'}} = {(3 y + 2)}^{2} - 9 (y^{2} + 4) = 9 y^{2} + 12 y + 4 - 9 y^{2} - 36

= 12 y - 32 = 4 (3 y - 8) t h e e q u . h a v e r o o t s \Leftrightarrow y ⩾ \frac{8}{3} i n t h i s

c a s e x_{1} = \frac{3 y + 2 + 2 \sqrt{3 y - 8}}{9} a n d x_{2} = \frac{3 y + 2 - 2 \sqrt{3 y - 8}}{9}

w e m u s t h a v e 3 x - y ⩽ 0 a n d x ⩾ 1

3 x_{1} - y = \frac{3 y + 2 + 2 \sqrt{3 y - 8} - 3 y}{3} = \frac{2 + 2 \sqrt{3 y - 8}}{3} > 0 (x_{1} t o

e l i m i n a t e) s o f^{- 1} (x) = \frac{1}{9} (3 x + 2 - 2 \sqrt{3 x - 8}) a n d f o r

x > \frac{3}{8} {(f^{- 1})}^{^{'}} (x) = \frac{1}{9} (3 - 2 \frac{3}{2 \sqrt{3 x - 8}}) \Rightarrow

{(f^{- 1})}^{^{'}} (x) = \frac{1}{3} (1 - \frac{1}{\sqrt{3 x - 8}}) .